Question

What is the pH of a solution that contains 50.00 mL of 0.10 M NH₃ to which 30.00 mL of 0.10 M HCl has been added?

Answer 1

consideration of this reaction:

NH₃ + HCl <===> NH₄Cl

Determine moles of NH₃ and HCl before mixing:

NH₃: (0.10mol/L) (0.0050 L) = 0.0005 mol
HCl: (0.10 mol/L) (0.0030 L) = 0.0003 mol

Determine moles of ammonia and ammonium ion after mixing:

NH₃: 0.0005 mol - 0.0003 mol = 0.0002 mol NH₃ remaining
NH₄⁺:0.0003 mol produced

Use the H-H Equation:

pH = pK_a + log (base/acid)

pH = 9.248 + log [(0.0002/0.080) / (0.0003/0.080)]

pH = 9.072

Note that the pK_a was used, NOT the pK_b Note also that the addition of acid caused the pH of the (now) buffer to decrease, that is to become more acidic. This is what we expect to happen. Lastly, note the inclusion of the new volume, 0.080 L. Both these values cancel out, so the new volume is rather inconsequential